同一張桌子上的多個左連接

Question

我有兩張桌子。 第一表=>構件{member_id，名稱，活性} 第二表=>積蓄{savings_id，member_id，月，年，數量，種類，支付}

成員表

+-----------+--------+--------+ 
| member_id | name | active | 
+-----------+--------+--------+ 
|  105 | Andri | 1  | 
|  106 | Steve | 1  | 
|  110 | Soraya | 1  | 
|  111 | Eva | 1  | 
|  112 | Sonia | 1  | 
+-----------+--------+--------+ 

儲蓄表

+------------+-----------+-------+------+--------+------+------+ 
| savings_id | member_id | month | year | amount | type | paid | 
+------------+-----------+-------+------+--------+------+------+ 
|   1 |  120 | NULL | NULL | 150000 | 1 | 1 | 
|   14 |  105 |  7 | 2014 | 80000 | 2 | 1 | 
|   15 |  105 |  7 | 2014 | 25000 | 3 | 1 | 
|   16 |  105 |  7 | 2014 | 60000 | 4 | 1 | 
|   17 |  105 |  7 | 2014 | 100000 | 5 | 1 | 
|   18 |  106 |  7 | 2014 | 80000 | 2 | 1 | 
|   19 |  106 |  7 | 2014 | 25000 | 3 | 1 | 
|   20 |  106 |  7 | 2014 | 60000 | 4 | 1 | 
|   21 |  106 |  7 | 2014 | 100000 | 5 | 1 | 
|   31 |  110 |  7 | 2014 | 25000 | 3 | 1 | 
|   32 |  110 |  7 | 2014 | 60000 | 4 | 1 | 
|   33 |  110 |  7 | 2014 | 100000 | 5 | 1 | 
|   34 |  111 |  7 | 2014 | 80000 | 2 | 1 | 
|   35 |  111 |  7 | 2014 | 25000 | 3 | 1 | 
|   36 |  111 |  7 | 2014 | 60000 | 4 | 1 | 
|   37 |  111 |  7 | 2014 | 100000 | 5 | 1 | 
|   38 |  112 |  7 | 2014 | 80000 | 2 | 1 | 
|   39 |  112 |  7 | 2014 | 25000 | 3 | 1 | 
|   40 |  112 |  7 | 2014 | 60000 | 4 | 1 | 
|   41 |  112 |  7 | 2014 | 100000 | 5 | 1 | 
|   85 |  105 |  7 | 2015 | 80000 | 2 | 1 | 
|   86 |  105 |  7 | 2015 | 25000 | 3 | 1 | 
|   87 |  105 |  7 | 2015 | 60000 | 4 | 1 | 
|   88 |  105 |  7 | 2015 | 100000 | 5 | 1 | 
|   89 |  106 |  7 | 2015 | 80000 | 2 |  | 
|   90 |  106 |  7 | 2015 | 25000 | 3 |  | 
|   91 |  106 |  7 | 2015 | 60000 | 4 |  | 
|   92 |  106 |  7 | 2015 | 100000 | 5 |  | 
|  101 |  110 |  7 | 2015 | 80000 | 2 |  | 
|  102 |  110 |  7 | 2015 | 25000 | 3 |  | 
|  103 |  110 |  7 | 2015 | 60000 | 4 |  | 
|  104 |  110 |  7 | 2015 | 100000 | 5 |  | 
|  105 |  111 |  7 | 2015 | 80000 | 2 | 1 | 
|  106 |  111 |  7 | 2015 | 25000 | 3 | 1 | 
|  107 |  111 |  7 | 2015 | 60000 | 4 | 1 | 
|  108 |  111 |  7 | 2015 | 100000 | 5 | 1 | 
|  109 |  112 |  7 | 2015 | 80000 | 2 |  | 
|  110 |  112 |  7 | 2015 | 25000 | 3 |  | 
|  111 |  112 |  7 | 2015 | 60000 | 4 |  | 
|  144 |  110 |  7 | 2014 | 50000 | 1 | 1 | 
+------------+-----------+-------+------+--------+------+------+ 

當會員儲蓄時，他們可以選擇5種儲蓄類型，我想要做的是將會員名單和他們所有的儲蓄。

這是MySQL查詢

SELECT m.member_id, name, 
SUM(s1.amount) as savings1, 
SUM(s2.amount) as savings2, 
SUM(s3.amount) as savings3, 
SUM(s4.amount) as savings4, 
SUM(s5.amount) as savings5 
FROM members m 
LEFT JOIN savings s1 ON s1.member_id = m.member_id AND s1.type = 1 AND s1.paid = 1 
LEFT JOIN savings s2 ON s2.member_id = m.member_id AND s2.type = 2 AND s2.paid = 1 
LEFT JOIN savings s3 ON s3.member_id = m.member_id AND s3.type = 3 AND s3.paid = 1 
LEFT JOIN savings s4 ON s4.member_id = m.member_id AND s4.type = 4 AND s4.paid = 1 
LEFT JOIN savings s5 ON s5.member_id = m.member_id AND s5.type = 5 AND s5.paid = 1 
WHERE 
active = 1 
GROUP BY m.member_id 

這是輸出

+-----------+--------+----------+----------+----------+----------+----------+ 
| member_id | name | savings1 | savings2 | savings3 | savings4 | savings5 | 
+-----------+--------+----------+----------+----------+----------+----------+ 
|  105 | Andri |  NULL | 1280000 | 400000 | 960000 | 1600000 | 
|  106 | Steve |  NULL | 80000 | 25000 | 60000 | 100000 | 
|  110 | Soraya | 50000 |  NULL | 25000 | 60000 | 100000 | 
|  111 | Eva |  NULL | 1280000 | 400000 | 960000 | 1600000 | 
|  112 | Sonia |  NULL | 80000 | 25000 | 60000 | 100000 | 
+-----------+--------+----------+----------+----------+----------+----------+ 

正如你可以看到的計算是不正確的，例如用於savings2部件105應該是160K。任何建議什麼應該是這個案件的查詢。

http://sqlfiddle.com/#!2/9eca9/1

Answer 1

問題是你正在形成完整的產品加盟在總結之前。因此，如果您的多個表中有多於一行，則最終會出現重複。如果你perform the join without summation，你可以清楚地看到發生了什麼。有兩種方法可以解決這個問題。

1. 是否所有的總和在派生表：

   SELECT m.member_id, name, 
   s1.amount as savings1, 
   s2.amount as savings2, 
   ... 
   FROM members m 
   LEFT JOIN (
       select SUM(amount) as amount, member_id 
       from savings 
       where type = 1 and paid = 1 
       group by member_id 
   ) s1 ON s1.member_id = m.member_id 
   LEFT JOIN (
       select SUM(amount) as amount, member_id 
       from savings 
       where type = 2 and paid = 1 
       group by member_id 
   ) s2 ON s2.member_id = m.member_id 
   ... 
   WHERE active = 1 
   GROUP BY m.member_id 
   

2. 加入一次，使用條件的總和：

   SELECT m.member_id, name, 
       SUM(CASE WHEN s.type = 1 then s.amount ELSE NULL END) as savings1, 
       SUM(CASE WHEN s.type = 2 then s.amount ELSE NULL END) as savings2, 
       ... 
   LEFT JOIN savings s s2 ON s.member_id = m.member_id AND s.paid = 1 
   WHERE active = 1 
   GROUP BY m.member_id 
   

Answer 2

你也許並不需要多個左連接和可以做的

SELECT 
m.member_id, 
m.name, 
SUM(case when s.type= 1 then s.amount end) as savings1, 
SUM(case when s.type= 2 then s.amount end) as savings2, 
SUM(case when s.type= 3 then s.amount end) as savings3, 
SUM(case when s.type= 4 then s.amount end) as savings4, 
SUM(case when s.type= 5 then s.amount end) as savings5 
FROM savings s 
join members m on m.member_id = s.member_id 
WHERE 
m.active = 1 
GROUP BY m.member_id 

Answer 3

這應該工作

SELECT m.member_id, name, 

sum((case when s1.type=1 then s1.amount end)) as savings1, 

sum((case when s1.type=2 then s1.amount end)) as savings2, 

sum((case when s1.type=3 then s1.amount end)) as savings3, 

sum((case when s1.type=4 then s1.amount end)) as savings4, 

sum((case when s1.type=5 then s1.amount end)) as savings5 

FROM members m 

LEFT JOIN savings s1 ON s1.member_id = m.member_id 

WHERE active = 1 and s1.paid=1 

GROUP BY m.member_id 

Answer 4

似乎加入有問題.. u能去上面的答案都是不錯的選擇..

U可以使用透視查詢也

SELECT m.member_id,s1.type, 
SUM(s1.amount) as savings 
FROM members m 
LEFT JOIN savings s1 ON s1.member_id = m.member_id AND s1.paid = 1 
WHERE active = 1 
GROUP BY m.member_id ,s1.type 

//樞軸這裏

，然後嘗試用於旋轉傢伙..這也將是不錯...